Find the sum of the n terms of the series whose nth term are given below : (i) `3n^(2)+2n` (ii) `2n^(3)+4n+1`
(iii) `2^(n)+3^(n)` (iv) `3^(n)+n^(3)`

Explanation

To find the sum of the n terms of the series given by their nth terms, we will solve each part step by step.

### (i) nth term: \(3n^2 + 2n\)

1. **Write the sum of the first n terms:**
\[ S_n = \sum_{k=1}^{n} (3k^2 + 2k) \]
This can be separated into two summations:
\[ S_n = 3 \sum_{k=1}^{n} k^2 + 2 \sum_{k=1}^{n} k \]

2. **Use the formulas for the summations:**
- The formula for the sum of the first n squares is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
- The formula for the sum of the first n natural numbers is:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]

3. **Substitute the formulas into the expression for \(S_n\):**
\[ S_n = 3 \left(\frac{n(n + 1)(2n + 1)}{6}\right) + 2 \left(\frac{n(n + 1)}{2}\right) \]

4. **Simplify the expression:**
\[ S_n = \frac{n(n + 1)(2n + 1)}{2} + n(n + 1) \]
\[ S_n = \frac{n(n + 1)(2n + 1 + 2)}{2} \]
\[ S_n = \frac{n(n + 1)(2n + 3)}{2} \]

### (ii) nth term: \(2n^3 + 4n + 1\)

1. **Write the sum of the first n terms:**
\[ S_n = \sum_{k=1}^{n} (2k^3 + 4k + 1) \]
This can be separated into three summations:
\[ S_n = 2 \sum_{k=1}^{n} k^3 + 4 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

2. **Use the formulas for the summations:**
- The formula for the sum of the first n cubes is:
\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n + 1)}{2}\right)^2 \]
- The sum of the first n natural numbers is:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
- The sum of 1 added n times is simply \(n\).

3. **Substitute the formulas into the expression for \(S_n\):**
\[ S_n = 2 \left(\frac{n(n + 1)}{2}\right)^2 + 4 \left(\frac{n(n + 1)}{2}\right) + n \]

4. **Simplify the expression:**
\[ S_n = \frac{n^2(n + 1)^2}{2} + 2n(n + 1) + n \]
\[ S_n = \frac{n(n + 1)}{2} \left(n(n + 1) + 4(n + 1) + 2\right) \]
\[ S_n = \frac{n(n + 1)}{2} (n^2 + 5n + 6) \]

### (iii) nth term: \(2^n + 3^n\)

1. **Write the sum of the first n terms:**
\[ S_n = \sum_{k=1}^{n} (2^k + 3^k) \]
This can be separated into two summations:
\[ S_n = \sum_{k=1}^{n} 2^k + \sum_{k=1}^{n} 3^k \]

2. **Use the formula for the sum of a geometric series:**
- The formula for the sum of a geometric series is:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
- For \(2^k\):
\[ \sum_{k=1}^{n} 2^k = 2 \frac{2^n - 1}{2 - 1} = 2(2^n - 1) \]
- For \(3^k\):
\[ \sum_{k=1}^{n} 3^k = 3 \frac{3^n - 1}{3 - 1} = \frac{3}{2}(3^n - 1) \]

3. **Combine the results:**
\[ S_n = 2(2^n - 1) + \frac{3}{2}(3^n - 1) \]

4. **Simplify the expression:**
\[ S_n = 2^{n + 1} - 2 + \frac{3}{2}(3^n - 1) \]
\[ S_n = 2^{n + 1} - 2 + \frac{3}{2}3^n - \frac{3}{2} \]
\[ S_n = 2^{n + 1} + \frac{3}{2}3^n - \frac{7}{2} \]

### (iv) nth term: \(3^n + n^3\)

1. **Write the sum of the first n terms:**
\[ S_n = \sum_{k=1}^{n} (3^k + k^3) \]
This can be separated into two summations:
\[ S_n = \sum_{k=1}^{n} 3^k + \sum_{k=1}^{n} k^3 \]

2. **Use the formula for the summations:**
- For \(3^k\):
\[ \sum_{k=1}^{n} 3^k = 3 \frac{3^n - 1}{3 - 1} = \frac{3}{2}(3^n - 1) \]
- For \(k^3\):
\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n + 1)}{2}\right)^2 \]

3. **Combine the results:**
\[ S_n = \frac{3}{2}(3^n - 1) + \left(\frac{n(n + 1)}{2}\right)^2 \]

4. **Simplify the expression:**
\[ S_n = \frac{3}{2}3^n - \frac{3}{2} + \frac{n^2(n + 1)^2}{4} \]

### Summary of Results:
1. \( S_n = \frac{n(n + 1)(2n + 3)}{2} \)
2. \( S_n = \frac{n(n + 1)(n^2 + 5n + 6)}{4} \)
3. \( S_n = 2^{n + 1} + \frac{3}{2}3^n - \frac{7}{2} \)
4. \( S_n = \frac{3}{2}(3^n - 1) + \left(\frac{n(n + 1)}{2}\right)^2 \)